A block of mass `10 kg` is moving in x-direction with a constant speed of `10 m//s`. it is subjected to a retarding force `F=-0.1 x J//m`. During its travel from `x =20m` to `x =30 m`. Its final kinetic energy will be.

Options

A. `450J`
B. `275J`
C. `250J`
D. `475 J`

Answer

D

Explanation

Here, `m=10kg, u=10 m//s, F=-0.1x J//m`
`[`as force is retarding `]`
`x_(1)=20cm, x_(2)=30cm`
As `W=int_(x_(1))^(x_(2))Fdx=int_(20)^(30)-0.1x dx =-0.1[(x^(2))/(2)]_(20)^(30)`
`=-(0.1)/(2)[(30)^(2)-(20)^(2)]=-(0.1)/(2)[900-400]`
`=-(0.1)/(2)[500]=-25J`
`K.E_(i)=(1)/(2)m u^(2)=(1)/(2)xx10xx(10)^(2)=500J`
According to work-energy theorem,
`W=K.E_(f)-K.E_(i)` or `-25=K.E_(f)-500`
`K.E_(f)=475J`